Problem: $\text D = \left[\begin{array}{rrr}-1 & 0 & 4 \\ 2 & 0 & 0\end{array}\right]$ and $\text F = \left[\begin{array}{rr}-1 & 1 \\ -1 & 3 \\ 2 & 4\end{array}\right]$ Let $\text {H = DF}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{D}$ and the first column of $\text{F}$. $ \text {H}=\left[\begin{array}{rr}{-1} & {0} & {4} \\ 2 & 0 & 0\end{array}\right]\left[\begin{array}{rr} {-1} & 1 \\ {-1} & 3 \\ {2} & 4\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(-1,0,4)\cdot(-1,-1,2)\\\\ &=-1 \cdot -1+0\cdot -1 + 4\cdot 2\\\\ &=9 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 \cdot 1+0\cdot 3 + 4\cdot 4 = 15$ (Choice B) B $2 \cdot -1+0\cdot -1 + 0\cdot 2 = -2$ (Choice C) C $0 \cdot -1+0\cdot 1 = 0$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}9 & 15 \\ -2 & 2\end{array}\right]$